# A triangle has sides A,B, and C. If the angle between sides A and B is pi/6, the angle between sides B and C is pi/12, and the length of B is 3, what is the area of the triangle?

May 21, 2018

The third angle of the given triangle is given by $\alpha = \frac{3}{4} \cdot \pi$ and the area is given by $A = a \cdot \frac{b}{2} \cdot \sin \left(\frac{\pi}{6}\right)$
The side length of $B C$ can be calculated with the theorem of sines.

#### Explanation:

$\alpha = \pi - \frac{\pi}{6} - \frac{\pi}{12} = \frac{3}{4} \cdot \pi$
$A = a \cdot \frac{b}{2} \cdot \sin \left(\frac{\pi}{6}\right)$
$\sin \frac{\frac{3}{4} \cdot \pi}{\sin} \left(\frac{\pi}{12}\right) = \frac{a}{3}$
so we get
$A = \frac{1}{2} \cdot 3 \cdot \sin \frac{\frac{3}{4} \cdot \pi}{\sin} \left(\frac{\pi}{12}\right) \cdot \sin \left(\frac{\pi}{6}\right)$
Further you can use:
$\sin \left(\frac{3}{4} \cdot \pi\right) = \frac{\sqrt{2}}{2}$
$\sin \left(\frac{\pi}{12}\right) = \frac{1}{4} \cdot \sqrt{2} \cdot \left(\sqrt{3} - 1\right)$
$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$